Question

show that the cube of any positive integer is of the form 4m, 4m+1 or 4m+3 for some integer m. ​

Answer 1

Step-by-step explanation:

Let a be the positive integer and b = 4.

Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.

So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3. (4q)³ = 64q³ = 4(16q³) = 4m, where m is some integer.

(4q + 1)³ = 64q³ + 48q² + 12q + 1 = 4(16q³+ 12q²+ 3) + 1 = 4m + 1, where m is some integer.

(4q + 2)³ = 64q³ + 96q² + 48q + 8 = 4(16q³ + 24q²+ 12q + 2) = 4m, where m is some integer.

(4q + 3)³ = 64q³ + 144q² + 108q + 27 = 4(16q³ + 36q² + 27q + 6) + 3 = 4m + 3,

Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.